Question

\(\frac{2}{x+1}\)-\(\frac{3}{x-1}\)=5

Answer 1

ĐKXĐ: \(x\ne-1;x\ne1\)

\(\frac{2}{x+1}-\frac{3}{x-1}=5\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{5\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow2x-2-3x-3=5x^2-5x+5x-5\)

\(\Leftrightarrow-x-5=5x^2-5\)

\(\Leftrightarrow-5x^2-x=0\)

\(\Leftrightarrow5x^2+x=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-\frac{1}{5}\left(TM\right)\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{0;-\frac{1}{5}\right\}\)